The ratio of energies of hydrogen atom in its first excited state to third excited state is

Options

A. `1/4`
B. `4/1`
C. `3/4`
D. `4/3`

Answer

To find the ratio of energies of a hydrogen atom in its first excited state to its third excited state, we can use the formula derived from Bohr's theory of the hydrogen atom. The energy of an electron in the nth energy level is given by:

\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step-by-step Solution:

1. **Identify the Energy Levels**:
- The first excited state corresponds to \( n = 2 \).
- The third excited state corresponds to \( n = 4 \).

2. **Calculate the Energy for Each State**:
- For the first excited state (\( n = 2 \)):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
- For the third excited state (\( n = 4 \)):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

3. **Find the Ratio of Energies**:
- The ratio of the energies \( \frac{E_2}{E_4} \) is:
\[ \frac{E_2}{E_4} = \frac{-3.4 \, \text{eV}}{-0.85 \, \text{eV}} = \frac{3.4}{0.85} \]
- Simplifying this gives:
\[ \frac{3.4}{0.85} = 4 \]

4. **Final Result**:
- Therefore, the ratio of the energies of the hydrogen atom in its first excited state to its third excited state is:
\[ \frac{E_2}{E_4} = 4 \]

### Conclusion:
The ratio of energies of the hydrogen atom in its first excited state to its third excited state is **4**.